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Question Number 120277 by bemath last updated on 30/Oct/20

lim_(x→∞) x^3 {(√(x^2 +(√(x^4 +1)))) − x(√2) } ?

$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{3}} \:\left\{\sqrt{{x}^{\mathrm{2}} +\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}\:−\:{x}\sqrt{\mathrm{2}}\:\right\}\:? \\ $$

Commented by benjo_mathlover last updated on 30/Oct/20

lim_(x→∞) x^4 {(√(1+(√(1+(1/x^4 ))))) −(√2) } setting (1/x) = z ∧ z→0 lim_(z→0) (((√(1+(√(1+z^4 ))))−(√2))/z^4 ) × (((√(1+(√(1+z^4 ))))+(√2))/( (√(1+(√(1+z^4 ))))+(√2))) lim_(z→0) (((√(1+z^4 ))−1)/z^4 ) × (1/(2(√2))) = ((√2)/4)×lim_(z→0) (z^4 /(z^4 ((√(1+z^4 ))+1))) = ((√2)/4) ×(1/2) = ((√2)/8)

$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{4}} \left\{\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}}\:−\sqrt{\mathrm{2}}\:\right\} \\ $$$$\:{setting}\:\frac{\mathrm{1}}{{x}}\:=\:{z}\:\wedge\:{z}\rightarrow\mathrm{0} \\ $$$$\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+{z}^{\mathrm{4}} }}−\sqrt{\mathrm{2}}}{{z}^{\mathrm{4}} }\:×\:\frac{\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+{z}^{\mathrm{4}} }}+\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+{z}^{\mathrm{4}} }}+\sqrt{\mathrm{2}}} \\ $$$$\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+{z}^{\mathrm{4}} }−\mathrm{1}}{{z}^{\mathrm{4}} }\:×\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}×\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{z}^{\mathrm{4}} }{{z}^{\mathrm{4}} \left(\sqrt{\mathrm{1}+{z}^{\mathrm{4}} }+\mathrm{1}\right)} \\ $$$$=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\:×\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{8}} \\ $$

Answered by Dwaipayan Shikari last updated on 30/Oct/20

x^3 ((√(x^2 +(√(x^4 +1))))−x(√2)) =x^3 ((√(x^2 +x^2 (√(1+(1/x^4 )))))−x(√2)) =x^3 (x(√(1+(√(1+(1/x^4 )))))−x(√2)) =x^4 ((√(1+1+(1/(2x^4 ))))−(√2)) =(√2)x^4 ((√(1+(1/(4x^4 ))))−1) =(√2) x^4 (1+(1/(8x^4 ))−1) =(1/(4(√2)))

$${x}^{\mathrm{3}} \left(\sqrt{{x}^{\mathrm{2}} +\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}−{x}\sqrt{\mathrm{2}}\right) \\ $$$$={x}^{\mathrm{3}} \left(\sqrt{{x}^{\mathrm{2}} +{x}^{\mathrm{2}} \sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}}−{x}\sqrt{\mathrm{2}}\right) \\ $$$$={x}^{\mathrm{3}} \left({x}\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}}−{x}\sqrt{\mathrm{2}}\right) \\ $$$$={x}^{\mathrm{4}} \left(\sqrt{\mathrm{1}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{4}} }}−\sqrt{\mathrm{2}}\right) \\ $$$$=\sqrt{\mathrm{2}}{x}^{\mathrm{4}} \left(\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{4}} }}−\mathrm{1}\right) \\ $$$$=\sqrt{\mathrm{2}}\:{x}^{\mathrm{4}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{8}{x}^{\mathrm{4}} }−\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}} \\ $$

Answered by bemath last updated on 30/Oct/20

lim_(x→∞) x^3 {((x^2 +(√(x^4 +1))−2x^2 )/( (√(x^2 +(√(x^4 +1)))) +x(√2))) } = lim_(x→∞) ((x^3 {(√(x^4 +1))−x^2 })/( (√(x^2 +(√(x^4 +1)))) +x(√2))) = lim_(x→∞) ((x^3 {x^4 +1−x^4 })/(x((√(1+(√(1+(1/x^4 )))))+(√2) )((√(x^4 +1))+x^2 ))) = lim_(x→∞) (x^3 /(x^3 ((√(1+(√(1+(1/x^4 )))))+(√2))((√(1+(1/x^4 )))+1)))= lim_(x→∞) (1/(((√(1+(√(1+(1/x^4 )))))+(√2))((√(1+(1/x^4 )))+1)))= (1/(((√2)+(√2))×2)) = (1/(4(√2)))

$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{3}} \:\left\{\frac{{x}^{\mathrm{2}} +\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}−\mathrm{2}{x}^{\mathrm{2}} }{\:\sqrt{{x}^{\mathrm{2}} +\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}\:+{x}\sqrt{\mathrm{2}}}\:\right\}\:= \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}^{\mathrm{3}} \:\left\{\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}−{x}^{\mathrm{2}} \:\right\}}{\:\sqrt{{x}^{\mathrm{2}} +\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}\:+{x}\sqrt{\mathrm{2}}}\:=\: \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}^{\mathrm{3}} \:\left\{{x}^{\mathrm{4}} +\mathrm{1}−{x}^{\mathrm{4}} \right\}}{{x}\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}}+\sqrt{\mathrm{2}}\:\right)\left(\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}+{x}^{\mathrm{2}} \:\right)}\:= \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{3}} \left(\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}}+\sqrt{\mathrm{2}}\right)\left(\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}+\mathrm{1}\right)}= \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}}+\sqrt{\mathrm{2}}\right)\left(\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}+\mathrm{1}\right)}= \\ $$$$\:\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}\right)×\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\: \\ $$$$\: \\ $$

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